Draw all the structural and geometrical (cis-trans) isomers of bromochloropropene.

Short Answer

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The structural and geometrical isomers of bromochloropropene are: 1. 1-bromo-2-chloropropene a. Cis-isomer (Cis-1-bromo-2-chloropropene) b. Trans-isomer (Trans-1-bromo-2-chloropropene) 2. 1-bromo-3-chloropropene

Step by step solution

01

Find the molecular formula of bromochloropropene

First, let's find out the molecular formula of bromochloropropene. Propene is an alkene with 3 carbon atoms and its formula is C3H6, adding bromine and chlorine we get the molecular formula: C3H5BrCl.
02

Draw the structural isomers

Since the parent chain is a propene, there are two possible positions for the chlorine and bromine atoms on the carbon chain. The structural isomers are as follows: 1. Chlorine on C1 and Bromine on C2(C1=C2-C3): 1-bromo-2-chloropropene 2. Chlorine on C1 and Bromine on C3(C1=C2-C3): 1-bromo-3-chloropropene
03

Draw the cis-trans isomers

For each structural isomer, we will now draw the geometrical isomers (cis-trans isomers) if any. 1. 1-bromo-2-chloropropene: Since the double bond is between C1 and C2, we can have different geometric arrangements for the bromine and chlorine atoms around the double bond, resulting in two cis-trans isomers. Cis-isomer: Bromine and Chlorine are on the same side of the double bond. Trans-isomer: Bromine and Chlorine are on opposite sides of the double bond. 2. 1-bromo-3-chloropropene: In this case, bromine is bonded to the sp2 carbon of the double bond, and chlorine is bonded to the sp3 carbon. Since there is a free rotation around single bond (C2-C3), we don't have any cis-trans isomers for this structure. Now we have all the structural and geometrical isomers of bromochloropropene: 1. 1-bromo-2-chloropropene a. Cis-isomer (Cis-1-bromo-2-chloropropene) b. Trans-isomer (Trans-1-bromo-2-chloropropene) 2. 1-bromo-3-chloropropene

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Most popular questions from this chapter

Cis-trans isomerism is also possible in molecules with rings. Draw the cis and trans isomers of 1,2 -dimethylcyclohexane. In Exercise 41, you drew all of the noncyclic structural and geometric isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~F}\). Now draw the cyclic structural and geometric isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~F}\).

Consider the following reactions. For parts \(\mathrm{b}-\mathrm{d}\), see Exercise \(62 .\) a. When \(\mathrm{C}_{5} \mathrm{H}_{12}\) is reacted with \(\mathrm{Cl}_{2}(g)\) in the presence of ultraviolet light, four different monochlorination products form. What is the structure of \(\mathrm{C}_{5} \mathrm{H}_{12}\) in this reaction? b. When \(\mathrm{C}_{4} \mathrm{H}_{8}\) is reacted with \(\mathrm{H}_{2} \mathrm{O}\), a tertiary alcohol is produced as the major product. What is the structure of \(\mathrm{C}_{4} \mathrm{H}_{8}\) in this reaction? c. When \(\mathrm{C}_{7} \mathrm{H}_{12}\) is reacted with \(\mathrm{HCl}\), 1 -chloro-1-methylcyclohexane is produced as the major product. What are the two possible structures for \(\mathrm{C}_{7} \mathrm{H}_{12}\) in this reaction? d. When a hydrocarbon is reacted with \(\mathrm{H}_{2} \mathrm{O}\) and the major product of this reaction is then oxidized, acetone (2-propanone) is produced. What is the structure of the hydrocarbon in this reaction? e. When \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\) is oxidized, a carboxylic acid is produced. What are the possible structures for \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\) in this reaction?

Draw the structures of the two dipeptides that can be formed from serine and alanine.

In glycine, the carboxylic acid group has \(K_{\mathrm{a}}=4.3 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=6.0 \times 10^{-5}\). Use these equilibrium constant values to calculate the equilibrium constants for the following. a. \({ }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) b. \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{OH}^{-}\) c. \({ }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}\)

Reagents such as \(\mathrm{HCl}, \mathrm{HBr}\), and \(\mathrm{HOH}\left(\mathrm{H}_{2} \mathrm{O}\right)\) can add across carbon- carbon double and triple bonds, with \(\mathrm{H}\) forming a bond to one of the carbon atoms in the multiple bond and \(\mathrm{Cl}, \mathrm{Br}\), or OH forming a bond to the other carbon atom in the multiple bond. In some cases, two products are possible. For the major organic product, the addition occurs so that the hydrogen atom in the reagent attaches to the carbon atom in the multiple bond that already has the greater number of hydrogen atoms bonded to it. With this rule in mind, draw the structure of the major product in each of the following reactions. a. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{*}}{\longrightarrow}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \longrightarrow\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}+2 \mathrm{HBr} \longrightarrow\) d. CC1=CCCC1 e. Cl CC=C(C)CC

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